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44
18. November 8th: Dedekind cuts.
18.1
.
Prove that the relation
≤
on
D
defned in Defnition 18.2 is antisymmetric.
Answer.
Assume (
A
1

B
1
)
≤
(
A
2

B
2
) and (
A
2

B
2
)
≤
(
A
1

B
1
). By defnition, this
means
A
1
⊆
A
2
and
A
2
⊆
A
1
. Since
⊆
is antisymmetric, this implies
A
1
=
A
2
. It
then Follows that
B
2
=
Q±
A
2
=
A
1
=
B
1
, proving that (
A
1

B
1
) = (
A
2

B
2
).
±
18.2
.
With notation as in the prooF oF Theorem 18.4, prove that
x<y
For all
x
∈
L
and all
y
∈
M
.
Answer.
Since
x
∈
L
(and by defnition oF union over Families, Defnition 5.7)
there exists an index
j
∈
I
such that
x
∈
A
j
. Since
y
∈
M
,
y
∈
B
i
For all
i
∈
I
(by
defnition oF intersection over Families), and in particular
y
∈
B
j
. Since (
A
j

B
j
) is a
Dedekind cut, it Follows that
x < y
, as required.
±
18.3
.
Provide a sensible defnition oF the ‘sum’ (
A
1

B
1
)+(
A
2

B
2
) oF two arbitrary
Dedekind cuts (
A
1

B
1
), (
A
2

B
2
). This should be done in such a way that iF
q
1
,q
2
∈
Q
and
ι
(
q
1
) = (
A
1

B
1
),
ι
(
q
2
)=(
A
2

B
2
), then (
A
1

B
1
A
2

B
2
)=
ι
(
q
1
+
q
2
).
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This note was uploaded on 12/14/2011 for the course MGF 3301 taught by Professor Aluffi during the Fall '11 term at FSU.
 Fall '11
 Aluffi

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